Understanding Monty Hall problem

Question

I was reading the Wikipedia page for the solution to Monty Hall problem using Bayes' theorem. I am very confused about this part

P(H3|X1) = 1/2 because this expression only depends on X1, not on any Ci. So, in this particular expression, the choosing of the host does not depend on where the car is, and there's only two remaining doors once X1 is chosen (for instance, P(H1|X1) = 0); and P(Ci,Xi) = P(Ci)P(Xi) because Ci and Xi are independent events (the player does not know where the car is in order to make a choice).

I still don't understand why P(H3|X1) = 1/2. I understand P(H3|X1) as the probability that the host will open door 3 given that the player chooses door 1 (and the car is inside door 1). Does the probability of 1/2 indicate that the host will either open door 2 or door 3? But I thought H3 is the event the host will open door 3? What am I missing?

Answer 1

Let's start with very minimal assumptions -- nothing about how the host chooses a door:$$ P(C1)=P(C2)=P(C3)=\tfrac{1}{3}\\ P(C1,C2)=P(C1,C3)=P(C2,C3)=0\\ P(Xi|Cj)=P(Xi) \forall i, j\\ P(Hi|Xi)=0 \\ P(Hi|Ci)= 0 $$

Given the situation in the article, we want $P(C1|X1, H3)$ and $P(C2|X1, H3)$.

$$P(C1|X1, H3)=\frac{P(H3, X1|C1) P(C1) }{P(X1, H3)}\\ P(H3, X1|C1)=P(H3|X1,C1)P(X1|C1)\\ P(X1, H3)=P(H3|X1)P(X1)\\ P(C1|X1, H3)=\frac{P(H3|X1,C1)P(X1|C1)P(C1)}{P(H3|X1)P(X1)} $$

The same formulas apply for $C2$. By our initial assumptions, we get some cancellation: $$ P(C1|X1, H3)=\frac{P(H3|X1,C1)\tfrac{1}{3}}{P(H3|X1)}\\ P(C2|X1, H3)=\frac{P(H3|X1,C2)\tfrac{1}{3}}{P(H3|X1)} $$ $P(H3|X1,C2)=1$ (host can't open the door with the car behind it), so:$$ P(C2|X1, H3)=\tfrac{1}{3}\frac{1}{P(H3|X1)} $$

Since $C1$, $C2$, and $C3$ are mutually exclusive and sum to 1, $P(H3|X1)=P(H3|X1,C1)P(C1)+P(H3|X1,C2)P(C2)+P(H3|X1,C3)P(C3)$

$P(H3|X1,C2)=1$ and $P(H3|X1,C3)=0$ so $P(H3|X1)=P(H3|X1,C1)P(C1)+P(C2)=\tfrac{1}{3}(P(H3|X1,C1)+1)$

If $P(H3|X1,C1)=\tfrac{1}{2}$, then $P(H3|X1)=\tfrac{1}{2}$ -- but only in that case. If the host hates door 3, and never chooses it unless they have to, $P(H3|X1,C1)=0$ and $P(H3|X1)=\tfrac{1}{3}$.

This makes for:$$ P(C2|X1, H3)=\frac{1}{(P(H3|X1,C1)+1)}\\ P(C1|X1, H3)=\tfrac{1}{3}\frac{P(H3|X1,C1)}{P(H3|X1)}=\frac{P(H3|X1,C1)}{(P(H3|X1,C1)+1)} $$

This looks like a pretty clear dependence on the host's strategy -- why?

Let's look at it from another angle: if the host and player were in cahoots, but couldn't communicate except through the game (but could coordinate in advance) -- could the host give the player a hint, through their choice of door?

Suppose the host always opens the lowest numbered door they can. In that case, we have:

	C1	C2	C3
X1	H2	H3	H2
X2	H3	H1	H1
X3	H2	H1	H1

If player picks 1, and the host responds 2 (X1, H3), the player would know that the car is behind 2 (C2) for sure.

Knowing nothing about the host's strategy, the player should use $1/2$ -- but that's not entirely satisfying -- because it seems like the host can influence the player's chances!

We can take a step back in the game, and look at the odds of the player winning, from the host's perspective. Call their final choice $Y$; take $X1$ as a given.

$$ P(Y1, X1, C1)+P(Y2, X1, C2)+P(Y3, X1, C3)=P(Y1, H2, X1, C1)+P(Y1, H3, X1, C1)+P(Y2, H3, X1, C2)+P(Y3, H2, X1, C3)=P(Y1| H2, X1, C1)P(H2, X1, C1)+P(Y1|H3, X1, C1)P(H3, X1, C1)+P(Y2|H3, X1, C2)P(H3, X1, C2)+P(Y3| H2, X1, C3)P(H2, X1, C3)\\ =P(Y1| H2, X1, C1)P(H2|X1, C1)P(X1,C1)+P(Y1|H3, X1, C1)P(H3| X1, C1)P(X1,C1)+P(Y2|H3, X1, C2)P(H3| X1, C2)P(X1,C2)+P(Y3| H2, X1, C3)P(H2| X1, C3)P(X1,C3) $$ Using $P(X1)=1$ and simplifying the cases where the host has no choice:$$ P(Y1, X1, C1)+P(Y2, X1, C2)+P(Y3, X1, C3)=P(Y1| H2, X1, C1)P(H2|X1, C1)\tfrac{1}{3}+P(Y1|H3, X1, C1)P(H3| X1, C1)\tfrac{1}{3}+P(Y2|H3, X1, C2)\tfrac{1}{3}+P(Y3| H2, X1, C3)\tfrac{1}{3} $$ $P(Y1| H2, X1, C1)$ and $P(Y1|H3, X1, C1)$ are both "keep the same door" for the player. Say the player does that with probability $x$. Then the other two terms are "remaining door," chosen with probability $1-x$.

Then: $$ P(Y1, X1, C1)+P(Y2, X1, C2)+P(Y3, X1, C3)=\tfrac{1}{3}(x (P(H2|X1, C1)+P(H3| X1, C1))+2(1-x)) $$

$P(H2|X1, C1)+P(H3| X1, C1)=1$ because the host has to choose one of H2 or H3. This leaves $\tfrac{1}{3}(2-x)$ as the odds of winning. This is maximized by $x=0$, regardless of how the host chooses a door.

Answer 2

Suppose the player chooses Door 1, and the prize is indeed behind Door 1. Which of the other two doors will Monty Hall open? Each has a probability of 1/2.

Answer 3

$P(H_3|X_1)$ does not have to be one half. The important thing about the Monty Hall problem is that the host knows which door holds the prize. If the contestant guesses wrong, which will happen two thirds of the time, Monty is left with only one door to open.

When the contestant guesses correctly, it doesn't Matter which door Monty opens. He could choose randomly, in which case $P(H_3|X_1)=\frac{1}{2}$ or he may have a policy that says he will choose the unselected door with the highest number in which case $P(H_3|X_1)=1$.

Worrying about how Monty chooses which door to open in case of a correct guess does not change the problem. The reason we have a two thirds chance of winning by adopting the switching strategy is because we win when we were originally wrong. Since we are originally wrong two thirds of the time we win two thirds of the time by switching.

To see this more clearly, imagine that the contestant chooses from ten doors. The contestant guesses door one and Monty opens doors three though ten. In this case there is only a one tenth chance of guessing correctly. Either way, Monty will choose all doors but one. If the contestant guesses wrong, he will open all doors except the one with the prize. If the contestant guesses correctly, he will be faced with the same problem as in the game with three doors. Once again, his strategy for choosing the open doors is irrelevant. The contestant will still win nine tenths of the time if he switches.

Answer 4

Use Bayes' theorem on it.

$$P(H_3|X_1) = \frac{P(X_1|H_3)P(H_3)}{P(X_1)}$$

$$= \frac{\frac12\frac13}{\frac13}$$

$$= \frac12$$

Answer 5

Using

we can construct three tables by collapsing/conditioning our model:

$$P(C_1) = 1$$

$$P(C_2) = 1$$

$$P(C_3) = 1$$

If we restore the sample space, we have the probabilities for all the entries $P(H|X)$ by scaling them with $P(C_1) = P(C_2) = P(C_3) = \frac{1}{3}$.

We have

$P(H_3|X_1) = P(H_3|X_1,C_1) + P(H_3|X_1,C_2) + P(H_3|X_1,C_3) = $
$\quad \quad \quad \quad \quad \quad 0.5 \cdot \frac{1}{3} + 1 \cdot \frac{1}{3} + 0 \cdot \frac{1}{3} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$

We are now in business and can crank out the solution to the following problem:

What is the probability the contestant wins the car if he initially selects door number $1$ but then switches when the host opens door $3$ (with a donkey)?

${\displaystyle {\begin{aligned}P(C2|H3,X1)&={\frac {P(C2,H3,X1)}{P(H3,X1)}}={\frac {P(H3|C2,X1)P(C2,X1)}{P(H3,X1)}}={\frac {P(C2)P(X1)}{P(H3|X1)P(X1)}}={\frac {1/3}{1/2}}={\frac {2}{3}}\end{aligned}}}$

Answer 6

Here is what turned the lightbulb on inside my head: Monte Hall is NOT going to open a door at random, all with equal probability. Instead, he's going to pick one of the doors you did not choose. Among those two, he will strategically take the one that's empty. Even if the probabilities were equal ex-ante, the subsequent non-random behavior of Hall makes the posterior probabilities unequal.

Answer 7

$P(H3|C1,X1)$ is the probability that the host opens door number three GIVEN that the guest picked door one (and the car was behind door one). The host never opens the door in which the car lies—otherwise it wouldn't be a very interesting game show—and never opens the door the guest picks.

Hence, since the door that the guest picks in this situation (door one) is the door to the car, the host eliminates only that door as an option, thus giving them two doors to choose from: Doors two and three. Assuming that the host has no inherent bias towards picking one number over the other, they would be expected to pick door three half of the time.

Answer 8

There are already a few answers, but none of them really stressed the point (in my opinion) which made me understand the Monty Hall Problem.

Once you chose a door, the doors you are offered to change to are never chosen at random since you are offered to change to another door after you made your initial decision. This seems like a minor detail, but this fact makes new information available, which results in a higher probability of winning if the doors are changed. Bayes' Theorem just formalizes this fact.

If the host selected a door prior to your choice, a potential offer to change the doors wouldn't alter the probabilities of winning. Hence the fact that the host offers you a new door is only relevant if this decision is made a posteriori; only in this case new information becomes available which can be used to increase the probability of winning.